The few steps before equation 3.20 in this paper imply that $$\exp\left(\sum_{n=1}^\infty{\frac{H_k^{(n)}}{n}x^n}\right)=\frac{\Gamma(1-x)\Gamma(k+1)}{\Gamma(k+1-x)}$$ Adjusting the starting point and then taking the limit $k\to\infty$, we have $$\exp\left(\sum_{n=2}^\infty{\frac{\zeta(n)}{n}x^n}\right)=e^{-\gamma x}\Gamma(1-x)$$ We can then say $$\sum_{n=2}^\infty{\frac{\zeta(n)}{n}x^n}=\sum_{n=0}^\infty{\frac{\frac{n!}{n+2}\zeta(n+2)}{n!}x^{n+2}}=\ln(e^{-\gamma x}\Gamma(1-x))=-\gamma x+\ln(\Gamma(1-x))$$ Therefore $$\sum_{n=0}^\infty{\frac{\frac{n!}{n+2}\zeta(n+2)}{n!}x^{n}}=\frac{-\gamma x+\ln(\Gamma(1-x))}{x^2}$$ And so we have $$\frac{n!}{n+2}\zeta(n+2)=\left.\frac{d^n}{dx^n}\frac{-\gamma x+\ln(\Gamma(1-x))}{x^2}\right|_{x=0}$$ I was hoping that we could use this to answer my main question:
Can we find the function $F$ where for some $b\in\mathbb{Z}$ $$\frac{n!}{n+2}\zeta(b(n+2))=\left.\frac{d^n}{dx^n}F(x)\right|_{x=0}$$